Termination w.r.t. Q proof of /tmp/tmp4W5lOP/Ex1

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

Q restricted rewrite system:
The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LENGTH(cons(X, L)) → LENGTH(L)
EQ(s(X), s(Y)) → EQ(X, Y)
TAKE(s(X), cons(Y, L)) → TAKE(X, L)
INF(X) → INF(s(X))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

LENGTH(cons(X, L)) → LENGTH(L)
EQ(s(X), s(Y)) → EQ(X, Y)
TAKE(s(X), cons(Y, L)) → TAKE(X, L)
INF(X) → INF(s(X))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LENGTH(cons(X, L)) → LENGTH(L)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LENGTH(cons(X, L)) → LENGTH(L)

R is empty.
The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LENGTH(cons(X, L)) → LENGTH(L)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

LENGTH(cons(X, L)) → LENGTH(L)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TAKE(s(X), cons(Y, L)) → TAKE(X, L)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TAKE(s(X), cons(Y, L)) → TAKE(X, L)

R is empty.
The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

TAKE(s(X), cons(Y, L)) → TAKE(X, L)

From the DPs we obtained the following set of size-change graphs:

TAKE(s(X), cons(Y, L)) → TAKE(X, L)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

INF(X) → INF(s(X))

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

INF(X) → INF(s(X))

R is empty.
The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

INF(X) → INF(s(X))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule INF(X) → INF(s(X)) we obtained the following new rules:

INF(s(z0)) → INF(s(s(z0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

INF(s(z0)) → INF(s(s(z0)))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule INF(s(z0)) → INF(s(s(z0))) we obtained the following new rules:

INF(s(s(z0))) → INF(s(s(s(z0))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ Instantiation

                      ↳ QDP

                        ↳ Instantiation

                          ↳ QDP

                            ↳ NonTerminationProof

          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

INF(s(s(z0))) → INF(s(s(s(z0))))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

INF(s(s(z0))) → INF(s(s(s(z0))))

The TRS R consists of the following rules:none

s = INF(s(s(z0))) evaluates to t =INF(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [z0 / s(z0)]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from INF(s(s(z0))) to INF(s(s(s(z0)))).

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

EQ(s(X), s(Y)) → EQ(X, Y)

The TRS R consists of the following rules:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

Q DP problem:
The TRS P consists of the following rules:

EQ(s(X), s(Y)) → EQ(X, Y)

R is empty.
The set Q consists of the following terms:

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

length(nil)
inf(x0)
eq(x0, x1)
take(s(x0), cons(x1, x2))
take(0, x0)
length(cons(x0, x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

EQ(s(X), s(Y)) → EQ(X, Y)

From the DPs we obtained the following set of size-change graphs:

EQ(s(X), s(Y)) → EQ(X, Y)
The graph contains the following edges 1 > 1, 2 > 2